1. http://lawrencekok.blogspot.com
Prepared by
Lawrence Kok
Video Tutorial on Energetics experiments and
Hess’s Law.

2. Average KE same
Heat (q) transfer thermal energy
from hot to cool due to temp diff
2
..
2
1
vmKE 
Average translational energy/KE
per particle
Heat Temperature
Heat vs Temperature
Symbol Q
Unit - Joule
Form of Energy Symbol T
Unit – C/K
Not Energy
At 80C
Distribution of molecular speed, Xe, Ar, Ne, He at same temp
2
.
2
1
vmKE 
80oC
Diff gas have same average KE per particle.
Click here Heat vs Temperature Click here specific heat capacity
He – mass low ↓ - speed v high ↑ Xe – mass high ↑ - speed v low ↓
Temp ᾳ Average KE
Higher temp - Higher average KE
2
..
2
1
vmKE 
Movement of particle, KE.
Heat energy
(energy in transfer)
80oC 50oC
degree of
hotness/coldness
Total KE/PE energy
of particles in motion
1 liter water 80C2 liter water 80C
Same Temp
Same Average kinetic energy per particle
Same Average speed
Same temp
Diff amt heat
More heat
energy
Less heat
energy
Heat energy
(energy in transfer)

3. Specific heat capacity
Amt heat needed to increase
temp of 1g of substance by 1C
Q = Heat transfer
Click here specific heat capacityClick here specific heat capacity
80oC 50oC
Warmer body
higher amt average KE
Energy transfer as heat
Gold
0.126J/g/K
Silver
0.90J/g/K
Water
4.18J/g/K
Cold body
lower amt average KE
Q = mcθ
Heat
Total KE/PE energy
of particles in motion
Symbol Q
Unit - Joule
Form of Energy
Amt heat energy Q, need to change temp depend
m = mass c = specific
heat capacity
θ = Temp diff
Lowest Highest
specific heat capacity
0.126J 4.18J
↓ ↓
to raise 1g to 1 K to raise 1g to 1K
Click here themochemistry notes
Coffee-cup calorimeter
constant pressure – no gas
Calorimetry - techniques used to measure
enthalpy changes during chemical processes.
Bomb calorimeter
Constant vol – gas released
80C
50C

4. Heat capacity bomb
Heat released
∆Hc is calculated.
Combustion - exo - temp water increase.
Specific heat capacity
Amount of heat needed to increase
temp of 1 g of substance by 1C
Q = Heat transfer
Q = mcθ
m = mass c = specific
heat capacity
θ = Temp diff
Coffee-cup calorimeter
constant pressure – no gas
Calorimetry - techniques used to measure
enthalpy changes during chemical processes.
Bomb calorimeter
Constant vol – gas released
Cup calorimeter
Determine specific heat capacity of X
m = 1000 g
Heated 200 C
5000 ml water
m = 5000g
c = 4.18
Ti = 20 C
Tf = 21.8 C
Heat lost by X = Heat gain by water
mc∆T = mc∆T
X
1000 x c (200 – 21.8) = 5000 x 4.18 x (21.8 – 20)
c = 37620/ 178200
c = 0.211J/g/K
Benzoic acid – used std – combustion 1g release 26.38 kJ
Combustion 0.579 g benzoic acid cause a 2.08°C increase in temp.
1.732g glucose combusted, temp increase of 3.64°C. Cal ΔHcomb glucose.
Bomb calorimeter (combustion)
Find heat capacity of bomb and ∆Hc X
Bomb sealed, fill with O2.
1g – 26.38kJ
0.579g – 26.38 x 0.579
Q = - 15.3kJ
∆Hc glucose = Qbomb
Find heat capacity bomb
Q bomb = c∆T
KkJc
T
Q
c
TcQ
/34.7
08.2
3.15


 Qbomb = c∆T
= 7.34 x 3.64
= 26.7 kJ
Insulated with water.
Combustion X
Find Q using
benzoic acid
1.732g – 26.7 kJ
180g – 2.78 x 103 kJmol-1
Click here bomb calorimeter
X
X
1. 2. 3.

5. System – rxn vessel
(rxn take place)
open system closed system isolated system
Enthalpy – Heat content/Amt heat energy in substance /KE + PE
- Energy stored as chemical bond + intermolecular force as potential energy
Exchange
energy
Exchange
matter
Exchange
energy
NO Exchange
energy
NO Exchange
matter
Heat(q) transfer from system to surrounding
↓
Exothermic. ∆H < 0
↓
HOT
Surrounding – rest of universe
Heat(q) transfer to system from surrounding
↓
Endothermic. ∆H > 0
↓
COLD
H
Time
H
Time
Heat
energy
Heat
energy
∆H = + ve
∆H = - ve
∆H system = O
reactionsystem
surrounding
No heat loss from system
(isolated system)
∆Hrxn = Heat absorb water (mc∆θ)
∆Hrxn = mc∆θ
water
Enthalpy Change = Heat of rxn = -∆H
2Mg(s) + O2(g) →2MgO(s) ∆H = -1200kJ mol-1
Enthalpy/H
(heat content)
2Mg + O2
2MgO
∆H= -1200kJ mol-1
- Energy neither created nor destroyed - Converted from one form into another.
- Amt heat lost by system equal to amt heat gain by its surrounding.
- Total energy system plus its surrounding is constant, if energy is conserved.

6. change
Energy Flow to/from System
System – KE and PE energy – Internal Energy (E)
Change Internal energy, ∆E = E final – E initial
Energy transfer as HEAT or WORK
Lose energy to surrounding as heat or work
E = sum kinetic energy/motion of molecule, and potential
energy represented by chemical bond bet atom
∆E = q + w
∆E = Change internal
energy
q = heat
transfer
w = work done
by/on system
Thermodynamics
Study of work, heat and energy on a system
Change Internal energy, ∆E = E final – E initial
Energy transfer as HEAT or WORK
Gain energy from surrounding as heat or work
Heat add , q = + 100 J
Work done by gas, w = - 20 J
∆E = + 100 – 20 = + 80 J
Q = Heat gain
+ 100J
w = work done by system = -20 J
w = work done on system = +20 J
Q = Heat lost
- 100J
Heat lost , q = - 100 J
Work done on gas, w = + 20 J
∆E = - 100 + 20 = - 80 J
∆E universe = ∆E sys + ∆E surrounding = 0
System – KE and PE energy – Internal Energy (E)
Heat and work Heat only
Q = Heat gain
+ 100 J
No work – no gas produced
Heat add , q = + 100 J
No work done = 0
∆E = q + w
∆E = + 100 = + 100 J
Q = Heat lost
- 100J
Heat lost, q = - 100 J
No work done = 0
∆E = q + w
∆E = - 100 = - 100 J
No work – no gas produced
H = E + PV ∆H = ∆E + P∆V
Enthalpy change w = work done
by/on gas
1st Law Thermodynamics
∆E = q + w ∆E = q
∆E = q + w
+ q = sys gain heat
- q = sys lose heat
+ w = work done on sys
- w = work done by sys
∆E = q + w
Work done by gas
No gas – No work

7. change
Energy Flow to/from System
System – KE and PE energy – Internal Energy (E)
Change Internal energy, ∆E = E final – E initial
Energy transfer as HEAT or WORK
Lose energy to surrounding as heat or work
E = sum kinetic energy/motion of molecule, and potential
energy represented by chemical bond bet atom
∆E = q + w
∆E = Change internal
energy
q = heat
transfer
w = work done
by/on system
Thermodynamics
Study of work, heat and energy on a system
Change Internal energy, ∆E = E final – E initial
Energy transfer as HEAT or WORK
Gain energy from surrounding as heat or work
No work done by/on system
∆E = q + w w = 0
∆E = + q (heat flow into/out system)
∆H = ∆E = q (heat gain/lost)
∆E universe = ∆E sys + ∆E surrounding = 0
System – KE and PE energy – Internal Energy (E)
Heat only – Exothermic and Endothermic reaction
Q = Heat gain
+ 100 J
No work – no gas produced
Heat add , q = + 100 J
No work done = 0
∆E = q + w
∆E = + 100 J
∆E = ∆H = + 100 J
Q = Heat lost
- 100J
Heat lost, q = - 100 J
No work done = 0
∆E = q + w
∆E = - 100 J
∆E = ∆H = - 100J
No work – no gas produced
H = E + PV ∆H = ∆E + P∆V
Constant pressure
Enthalpy change w = work done
by/on gas
1st Law Thermodynamics
P∆V = 0
∆E = q + w∆H = ∆E + P∆V
∆E = q + 0
↓
∆E = q
No gas produced
V = 0
∆H = ∆E + 0
↓
∆H = ∆E
At constant pressure/no gas produced
∆H = q
∆Enthalpy change = Heat gain or lost
No work done
w = 0
H
E
E
∆H = + 100J
H ∆H = - 100J
Enthalpy Change

8. Heat(q) transfer from system to surrounding
↓
Exothermic ∆H < 0
↓
HOT
Heat
energy ∆H = - ve
Enthalpy Change = Heat of rxn = -∆H
Mg(s) + ½ O2(g) → MgO(s) ∆H = -1200kJ mol-1
Mg + ½ O2
MgO
∆H= -1200
- Energy neither created nor destroyed - Converted from one form into another.
- Amt heat lost by system equal to amt heat gain by its surrounding.
- Total energy system plus its surrounding is constant, if energy is conserved.
Reactant (Higher energy - Less stable/weaker bond)
Product (Lower energy - More stable/strong bond)
Temp surrounding ↑
Exothermic rxn
Combustion C + O2 → CO2
Neutralization H+ + OH- → H2O
Displacement Zn + CuSO4 → ZnSO4 + Cu
Condensation H2O(g) → H2O(l)
Freezing H2O(l) → H2O(s)
Precipitation Ag+ + CI- → AgCI(s)
Endothermic rxn
Dissolve NH4 salt NH4CI (s) → NH4
+ + CI -
Dissolve salt MgSO4. 7H2O(s) → MgSO4(aq
CuSO4. 5H2O(s) → CuSO4(aq)
Na2CO3.10H2O(s) → Na2CO3(aq
Evaporation/Boiling H2O(l) → H2O(g)
Melting H2O(s) → H2O(l)
Heat(q) transfer to system from surrounding
↓
Endothermic. ∆H > 0
↓
COLD
Heat
energy
Reactant (Lower energy - More stable/strong bond)
Product (Higher energy - Less stable/weak bond)
∆H = + ve Temp surrounding ↓
Click here thermodynamics
∆H= + 16
NH4CI (s)
NH4CI (aq)
Enthalpy Change = Heat of rxn = -∆H
NH4CI (s) → NH4CI (aq) ∆H = + 16 kJ mol-1
Click here enthalpy
3000
1800
116
100
E
X
O
E
N
D
O

9. NaCI (s)
Na(s) + ½CI2 (g))
LiCI (s)
Li+
(g) + CI–
(g)
AgCI
Ag+ + CI-
NaCI + H2O
HCI + NaOH
ZnSO4 + Cu
Zn + CuSO4
Li+
(aq)
Li+
(g) + H2O
LiCI(aq)
LiCI+ H2O
2CO2 + 3H2O
C2H5OH + 3O2
- Energy neither created nor destroyed - Converted from one form into another.
- Amt heat lost by system equal to amt heat gain by its surrounding.
- Total energy system plus its surrounding is constant, if energy is conserved.
Std Enthalpy Changes ∆Hθ
Std condition
Pressure
100kPa
Conc 1M All substance
at std states
Temp
298K
Bond Breaking
Heat energy absorbed – break bond
Bond Making
Heat energy released – make bond
Std ∆Hc
θ combustion
Std Enthalpy Changes ∆Hθ
∆H for complete combustion
1 mol sub in std state in excess O2
∆H when 1 mol solute
dissolved form infinitely dilute sol
Std ∆Hsol
θ solution
∆H when 1 mol
gaseous ion is hydrated
Std ∆Hhyd
θ hydration
∆H when 1 mol metal
is displaced from its sol
Std ∆Hd
θ displacement
∆H when 1 mol
H+ react OH- to form 1 mol H2O
C2H5OH + 3O2 → 2CO2 + 3H2O LiCI(s) + H2O → LiCI(aq)
Ag+ + CI - → AgCI (s)
Zn + CuSO4 → ZnSO4 + Cu
Std ∆Hlat
θ lattice
∆H when 1 mol
precipitate form from its ion
Std ∆Hppt
θ precipitationStd ∆Hn
θ neutralization
∆H when 1 mol crystalline
sub form from its gaseous ion
HCI + NaOH → NaCI + H2O Li+
(g) + CI–
(g) → LiCI (s)
Li+
(g) + H2O→ Li+
(aq)
∆H = - ve ∆H = - ve ∆H = - ve ∆H = - ve
∆H = - ve ∆H = - ve ∆H = - ve
∆H when 1 mol form from
its element under std condition
∆H = - ve
Std ∆Hf
θ formation
Na(s) + ½CI2 (g)→ NaCI (s)

10. ∆H = - 50 – (+12)
= - 62 kJ mol -1
∆H = - 50 kJ mol -1
Cold water = 50g
CuSO4 .100H2O
CuSO4 (s) + 95H2O
→ CuSO4 .100H2O
CuSO4 (s) + 100H2O
→ CuSO4 .100H2O
Mass cold water add = 50 g
Mass warm water add = 50 g
Initial Temp flask/cold water = 23.1 C
Initial Temp warm water = 41.3 C
Final Temp mixture = 31 C
CuSO4 (s) + 5H2O → CuSO4 .5H2O ∆H = ?
Slow rxn – heat lost huge – flask used.
Heat capacity flask must be determined.
CuSO4 (s) + 5H2O → CuSO4 .5H2O
1. Find heat capacity flask
Ti = 23.1C
Hot water = 50g
T i = 41.3C
No heat loss from system
(isolated system)
↓
∆H system = O
Heat lost hot H2O=Heat absorb cold H2O + Heat absorb flask
(mc∆θ) (mc∆θ) (c ∆θ)
50 x 4.18 x (41.3 – 31) = 50 x 4.18 x (31-23.1) + c x (31-23.1)
Water Flask CuSO4
Heat capacity flask, c = 63.5JK-1
Mass CuSO4 = 3.99 g (0.025mol)
Mass water = 45 g
T initial flask/water = 22.5 C
T final = 27.5 C
Mass CuSO4 5H2O = 6.24 g (0.025mol)
Mass water = 42.75 g
T initial flask/water = 23 C
T final = 21.8 C
2. Find ∆H CuSO4 +100H2O → CuSO4 .100H2O
3. Find ∆H CuSO4 .5H2O + 95H2O → CuSO4 .100H2O
Hess’s Law
CuSO4 + H2O → CuSO4 .100H2O CuSO4 .5H2O + H2O → CuSO4 .100H2O
Water Flask CuSO4 5H2O
∆Hrxn = Heat absorb water + vacuum
(mc∆θ) + (c∆θ)
∆Hrxn = Heat absorb water + vacuum
(mc∆θ) + (c∆θ)
∆Hrxn= 45 x 4.18 x 5 + 63.5 x 5
∆Hrxn= - 1.25 kJ
∆Hrxn= 42.75 x 4.18 x 1.2 + 63.5 x 1.2
∆Hrxn= + 0.299 kJ
0.025 mol = - 1.25 kJ
1 mol = - 50 kJ mol-1
0.025 mol = + 0.299 kJ
1 mol = + 12 kJ mol-1
CuSO4 (s) + 5H2O → CuSO4 .5H2O
CuSO4 .100H2O
∆H = + 12 kJ mol -1
Assumption
No heat lost from system ∆H = 0
Water has density = 1.0 gml-1
Sol diluted Vol CuSO4 = Vol H2O
All heat transfer to water + flask
Rxn slow – lose heat to surrounding
Plot Temp/time – extrapolation done, Temp correction
limiting
Enthalpy change ∆H - using calorimeter
without temp
correction
Lit value = - 78 kJ mol -1
CONTINUE

11. Time/
m
0 0.5 1 1.5 2 2.5 3 3.5 4
Temp/C 22 22 22 22 22 27 28 27 26
∆H = - 60 kJ mol -1
CuSO4 .100H2O
CuSO4 (s) + 95H2O
→ CuSO4 .100H2O
CuSO4 (s) + 100H2O
→ CuSO4 .100H2O
CuSO4 (s) + 5H2O → CuSO4 .5H2O ∆H = ?
CuSO4 (s) + 5H2O → CuSO4 .5H2O
Water Flask CuSO4
Mass CuSO4 = 3.99 g (0.025 mol)
Mass water = 45 g
T initial mix = 22 C
T final = 28 C
Mass CuSO4 5H2O = 6.24 g (0.025 mol)
Mass water = 42.75 g
T initial mix = 23 C
T final = 21 C
Find ∆H CuSO4 + 100H2O → CuSO4 .100H2O
Find ∆H CuSO4 .5H2O + 95H2O → CuSO4 .100H2O
Hess’s Law
CuSO4 + H2O → CuSO4 .100H2O CuSO4 .5H2O + H2O → CuSO4 .100H2O
Water Flask CuSO4 5H2O
∆Hrxn = Heat absorb water + flask
(mc∆θ) + (c∆θ)
∆Hrxn = Heat absorb water + flask
(mc∆θ) + (c∆θ)
∆Hrxn= 45 x 4.18 x 6 + 63.5 x 6
∆Hrxn= - 1.5 kJ
∆Hrxn= 42.75 x 4.18 x 2 + 63.5 x 2
∆Hrxn= + 0.48kJ
0.025 mol = - 1.5 kJ
1 mol = - 60 kJ mol-1
0.025 mol = + 0.48 kJ
1 mol = + 19 kJ mol-1
CuSO4 (s) + 5H2O → CuSO4 .5H2O
CuSO4 .100H2O
∆H = + 19 kJ mol -1
∆H = - 60 – (+19)
= - 69 kJ mol -1
Rxn slow – lose heat to surrounding
Plot Temp/time – extrapolation done, Temp correction
limiting
Enthalpy change ∆H using calorimeter
Data collection
Temp correction – using cooling curve for last 5 m
time, x = 2
initial Temp = 22 C
final Temp = 28 C
Extrapolation best curve fit
y = -2.68x + 33
y = -2.68 x 2 + 33
y = 28 (Max Temp)
Excel plot
CuSO4 + H2O → CuSO4 .100H2O
(Exothermic) Heat released
CuSO4 .5H2O + H2O → CuSO4 .100H2O
(Endothermic) – Heat absorbed
Temp correction – using warming curve for last 5 m
Time/
m
0 0.5 1 1.5 2 2.5 3 3.5 4
Temp/C 23 23 23 23 23 22 22 22.4 22.7
initial Temp = 23 C
time, x = 2
final Temp = 21 C
Excel plot
Extrapolation curve fit
y = + 0.8 x + 19.4
y = + 0.8 x 2 + 19.4
y = 21 (Min Temp)
Lit value = - 78 kJ mol -1
with temp
correction

12. ∆H displacement using calorimeter
Time/m 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
Temp/C 27 27 27 27 27 27 27 66 71.4 71.8 70
Error Analysis
Heat loss to surrounding
Heat capacity cup / thermometer ignored
Heat capacity sol not 4.18
Mass CuSO4 ignored
Excess Zn absorb heat
Cu form shield Zn from reacting
Zn + CuSO4 → ZnSO4 + Cu
No heat loss from system
(isolated system)
↓
∆H system = O
Heat lost rxn = Heat absorb CuSO4
= (mc∆θ)
= 25 x 4.18 x (73 – 27)
= 25 x 4.18 x 46
= - 4807J
V, CuSO4 = 25ml
M, = 1M
T initial = 27 C
T final = 73 C
Coffee cup calorimeter Data collection
Extrapolation best curve fit
y = -2.68x + 81.3
y = -2.68 x 3 + 81.3
y = 73 (Temp)
initial Temp = 27 C
final Temp = 73 C
time, x = 3
using Excel plot
Using data logger
Min gradient
Extrapolation min fit
y = -2.68x + 81.3
y = -2.68 x 3 + 81.3
y = 73 (Min Temp)
Extrapolation max fit
y = -3.22x + 85.5
y = -3.22 x 3 + 85.5
y = 76 (Max Temp)
Using data logger
Max gradient
Ave Temp = (76 + 73)/2
= 75 C
0.025 mol = - 4807 J
1 mol = - 192 kJ mol-1
Enthalpy displacement = 1 mol of Cu 2+ displaced
Amt Cu 2+ = MV
used = 1 x 0.025
= 0.025 mol
Zn + CuSO4 → ZnSO4 + Cu ∆H = -192 kJ mol -1
% Error = (217 – 192) x 100% = 11%
217
Lit value = - 217kJ mol -1
Assump valid if rate of cooling constant
If not constant - extrapolation is wrong
Rxn slow – lose heat to surrounding
Plot Temp/time – extrapolation done, Temp correction
Temp correction – using cooling curve for last 5 m
Measure temp CuSO4,
every 0.5m interval
then add zinc in excess
Zinc powder
(excess)
Min/Max gradient

13. ∆H = - 113 kJ mol -1
Enthalpy change ∆H using calorimeter
Cold water = 50 g
MgSO4 .100H2O
MgSO4.7H2O + 93H2O
→ MgSO4 .100H2O
MgSO4 + 100H2O
→ MgSO4 .100H2O
Mass cold water add = 50 g
Mass warm water add = 50 g
Initial Temp flask/cold water = 23.1 C
Initial Temp warm water = 41.3 C
Final Temp flask/mixture = 31 C
MgSO4 (s) + 7H2O → MgSO4 .7H2O ∆H = ?
Slow rxn – heat lost huge – flask used.
Heat capacity flask must be determined.
MgSO4(s) + 7H2O → MgSO4 .7H2O
1. Find heat capacity flask
Ti = 23.1 C
Hot water = 50 g
T i = 41.3 C
No heat loss from system
(isolated system)
↓
∆H system = O
Heat lost hot H2O=Heat absorb cold H2O + Heat absorb flask
(mc∆θ) (mc∆θ) (c ∆θ)
50 x 4.18 x (41.3 – 31) = 50 x 4.18 x (31-23.1) + c x (31-23.1)
Water Flask MgSO4
Heat capacity flask, c = 63.5JK-1
Mass MgSO4 = 3.01 g (0.025mol)
Mass water = 45 g
T initial mix = 24.1 C
T final = 35.4 C
Mass MgSO4 7H2O = 6.16 g (0.025mol)
Mass water = 41.8 g
T initial mix = 24.8 C
T final = 23.4 C
2. Find ∆H MgSO4 +100H2O → MgSO4 .100H2O
3. Find ∆H MgSO4 .7H2O + 93H2O → MgSO4 .100H2O
Hess’s Law
MgSO4 + H2O → MgSO4 .100H2O MgSO4 .7H2O + H2O → MgSO4 .100H2O
Water Flask MgSO4 .7H2O
∆Hrxn = Heat absorb water + flask
(mc∆θ) + (c∆θ)
∆Hrxn = Heat absorb water + flask
(mc∆θ) + (c∆θ)
∆Hrxn= 45 x 4.18 x 11.3 + 63.5 x 11.3
∆Hrxn= - 2.83 kJ
∆Hrxn= 41.8 x 4.18 x 1.4 + 63.5 x 1.4
∆Hrxn= + 0.3 kJ
0.025 mol = - 2.83 kJ
1 mol = - 113 kJ mol-1
0.025 mol = + 0.3 kJ
1 mol = + 12 kJ mol-1
MgSO4 (s) + 7H2O → MgSO4 .7H2O
MgSO4 .100H2O
∆H = + 12 kJ mol -1
∆H = - 113 - 12
= - 125 kJ mol -1
Assumption
No heat lost from system ∆H = 0
Water has density = 1.00g ml-1
Sol diluted Vol MgSO4 = Vol H2O
All heat transfer to water + flask
Rxn fast – heat lost to surrounding minimized
Dont need to Plot Temp/time
No extrapolation
Error Analysis
Heat loss to surrounding
Heat capacity sol is not 4.18
Mass MgSO4 ignored
Impurity present
MgSo4 already hydrated
limiting

14. ∆H = - 286 kJ mol -1
∆H = - 442 kJ mol -1
Enthalpy change ∆H using calorimeter
Cold water = 50g
Mass cold water add = 50g
Mass warm water add = 50g
Initial Temp flask/cold water = 23.1C
Initial Temp warm water = 41.3C
Final Temp flask/mixture = 31C
Mg(s) + ½ O2 → MgO ∆H = ?
Slow rxn – heat lost huge – flask is used.
Heat capacity flask must be determined.
Mg + ½ O2 → MgO
1. Find heat capacity flask
Ti = 23.1C
Hot water = 50g
T i = 41.3C
No heat loss from system
(isolated system)
↓
∆H system = O
Heat lost hot H2O=Heat absorb cold H2O + Heat absorb Flask
(mc∆θ) (mc∆θ) (c ∆θ)
50 x 4.18 x (41.3 – 31) = 50 x 4.18 x (31-23.1) + c x (31-23.1)
HCI Flask Mg
Heat capacity flask, c = 63.5JK-1
Mass Mg = 0.5 g (0.02 mol)
Vol/Conc HCI = 100 g, 0.1M
T initial mix = 22 C
T final = 41 C
Mass MgO = 1 g (o.o248 mol)
Vol/Conc HCI = 100 g, 0.1M
T initial mix = 22 C
T final = 28.4 C
2. Find ∆H Mg + 2HCI → MgCI2 + H2
3. Find ∆H MgO + 2HCI → MgCI2 + H2O
4. Find H2 + ½ O2 → H2O
Hess’s Law
∆H Mg + 2HCI → MgCI2 + H2
∆Hrxn = Heat absorb water + flask
(mc∆θ) + (c∆θ)
∆Hrxn = Heat absorb water + flask
(mc∆θ) + (c∆θ)
∆Hrxn= 100 x 4.18 x 19 + 63.5 x 19
∆Hrxn= - 9.11kJ
∆Hrxn= 100 x 4.18 x 6.4 + 63.5 x 6.4
∆Hrxn= -3.1 kJ
0.02 mol = - 9.11 kJ
1 mol = - 442 kJ mol-1
0.0248 mol = - 3.1 kJ
1 mol = - 125 kJ mol-1
∆H = - 125 kJ mol -1
∆H = - 442 – 286 + 125
= - 603 kJ mol -1
Assumption
No heat lost from system ∆H = 0
Water has density = 1.00g ml-1
Sol diluted Vol HCI = Vol H2O
All heat transfer to water + flask
Rxn fast – heat lost to surrounding minimized
Dont need to Plot Temp/time
No extrapolation
Error Analysis
Heat loss to surrounding
Heat capacity sol is not 4.18
Mass MgO ignored
Impurity present
Effervescence cause loss Mg
+ 2HCI
MgCI2 + H2
HCI Flask MgO
+ ½ O2
MgCI2 + H2O
+ 2HCI
MgCI2 + H2O
∆H MgO + 2HCI → MgCI2 + H2O
Mg + ½ O2 → MgO
MgCI2 + H2
+ 2HCI
MgCI2 + H2O
+ ½ O2
limiting
MgCI2 + H2O
Data given

15. 2KHCO3(s) → K2CO3 + CO2 + H2O
∆H = + 51.4 kJ mol -1
Enthalpy change ∆H for rxn using calorimeter
Cold water = 50g
Mass cold water add = 50 g
Mass warm water add = 50 g
Initial Temp flask/cold water = 23.1 C
Initial Temp warm water = 41.3 C
Final Temp flask/mixture = 31 C
2KHCO3(s) → K2CO3 + CO2 + H2O ∆H = ?
Slow rxn – heat lost huge – vacuum flask is used.
Heat capacity of flask must be determined.
1. Find heat capacity vacuum
Ti = 23.1C
Hot water = 50g
T i = 41.3C
No heat loss from system
(isolated system)
↓
∆H system = O
Heat lost hot H2O=Heat absorb cold H2O + Heat absorb flask
(mc∆θ) (mc∆θ) (c ∆θ)
50 x 4.18 x (41.3 – 31) = 50 x 4.18 x (31-23.1) + c x (31-23.1)
HCI Flask KHCO3
Heat capacity vacuum, c = 63.5JK-1
Mass KHCO3 = 3.5 g (0.035 mol)
Vol/Conc HCI = 30 g, 2M
T initial mix = 25 C
T final = 20 C
Mass K2CO3 = 2.75 g (0.02 mol)
Vol/Conc HCI = 30 g, 2M
T initial mix = 25 C
T final = 28 C
2. Find ∆H 2KHCO3 + 2HCI → 2KCI + 2CO2 + 2H2O
3. Find ∆H K2CO3 + 2HCI → 2KCI + CO2 + H2O
Hess’s Law
∆Hrxn = Heat absorb water + vacuum
(mc∆θ) + (c∆θ)
∆Hrxn = Heat absorb water + vacuum
(mc∆θ) + (c∆θ)
∆Hrxn= 30 x 4.18 x 5 + 63.5 x 5
∆Hrxn= + 0.9 kJ
∆Hrxn= 30 x 4.18 x 3 + 63.5 x 3
∆Hrxn= -0.56 kJ
0.035 mol = + 0.9 kJ
1 mol = + 25.7 kJ mol-1
0.02 mol = - 0.56 kJ
1 mol = - 28 kJ mol-1
∆H = - 28 kJ mol -1
∆H = +51.4 – (-28)
= + 79 kJ mol -1
Assumption
No heat lost from system ∆H = 0
Water has density = 1.00g ml-1
Sol diluted Vol HCI = Vol H2O
All heat transfer to water + vacuum
Rxn fast – heat lost to surrounding minimized
Dont need to Plot Temp/time
No extrapolation
Error Analysis
Heat loss to surrounding
Heat capacity sol is not 4.18
Mass of MgO ignored
Impurity present
Effervescence cause loss Mg
+ 2HCI
HCI Flask K2CO3
2KCI + 2CO2 + 2H2O
+ 2HCI
+ 2HCI
limiting
2KHCO3(s) → K2CO3 + CO2 + H2O
2KCI + CO2 + H2O
2KHCO3 + 2HCI → 2KCI + 2CO2 + 2H2O K2CO3 + 2HCI → 2KCI + CO2 + H2O
x 2
2KCI + 2CO2 + 2H2O 2KCI + CO2 + H2O
+ 2HCI

16. Vol/ml 0 5 10 15 20 25 30 35 40 45 50
Temp/C 22 24. 26 28 30 31 30 30 29 28 27
Error Analysis
Heat loss to surrounding
Heat capacity cup / thermometer ignored
Heat capacity solution is not 4.18
Mass of HCI/NaOH ignored
∆H neutralization rxn
HCI + NaOH → NaCI + H2O
2.M HCI – 5 ml added and temp recorded
∆H rxn = Heat absorb water
= (mc∆θ)
= 75 x 4.18 x (31 – 22)
= 75 x 4.18 x 9
= - 2821J
V, NaOH = 50 ml
M, = 1M
T initial = 22 C
T final = 31 C by extrapolation.
Coffee cup calorimeter Data collection
Vol, acid = 25
Max Temp = 31
∆T = ( 31 – 22) = 9 C
0.05 mol = - 2821 J
1 mol = - 56 kJ mol-1
Amt OH- = MV
used = 1 x 0.05
= 0.05 mol
% Error = (57.3 – 56) x 100% = 2 %
57.3
Lit value = - 57.3 kJ mol -1
Assump valid if rate of cooling constant
If not constant - extrapolation is wrong
Rxn slow – lose heat to surrounding
Plot Temp/vol – extrapolation done, Temp correction
Temp correction – using heating/cooling curve
Thermometric Titration
HCI + NaOH → NaCI + H2O ∆H = - 56 kJ mol -1
initial Temp
= 22 C
final Temp
= 31 C
Excel plot
Heating curve
y = 0.362x + 22.4
Cooling curve
y = - 0.16 x + 35.7
Heating curve
y = 0.362x + 22.4
Cooling curve
y = - 0.16 x + 35.7
Solving for , x and y
y = + 0.362 x + 22.4 y = - 0.16 x + 35.7
+0.362 x + 22.4 = -0.16 x + 35.7
x = 25 (vol acid) y = 31 (Temp)
Strong acid with Strong alkali
∆H neutralization = 1 mol H+ react 1 mol OH- form 1 mol H2O
Strong acid vs Strong alkali
Weak acid vs strong alkali

17. 0.05 mol = - 2541 J
1 mol = - 51 kJ mol-1
Vol/ml 0 5 10 15 20 25 30 35 40 45 50
Temp/C 21 22. 24 26 27 28 28 28 27 27 26
∆Hn always same (-57.3) regardless strong or weak acid
H+ + OH- → H2O ∆H = 57.3 kJmol-1
∆H n weak acid vs strong alkali is lower (less –ve)
Weak acid - molecule doesn’t dissociate completely
Heat absorb to ionize/dissociate molecule – less heat released
∆H neutralization rxn
CH3COOH + NaOH → CH3COONa + H2O
1.9M CH3COOH – 5 ml added and temp recorded
∆H rxn = Heat absorb water
= (mc∆θ)
= 76 x 4.18 x (29 – 21)
= 76 x 4.18 x 8
= - 2541J
V, NaOH = 50 ml
M, = 1M
T initial = 21 C
T final = 29 C
Coffee cup calorimeter Data collection
Vol, acid = 26
Max Temp = 29
∆T = (29– 21) = 8 C
∆H neutralization = 1 mol H+ react 1 mol OH- form 1 mol H2O
Amt OH- = MV
used = 1 x 0.05
= 0.05 mol
% Error = (57.3 – 51) x 100% = 11 %
57.3
Lit value = - 57.3 kJ mol -1
Assump valid if rate of cooling constant
If not constant - extrapolation is wrong
Rxn slow – lose heat to surrounding
Plot Temp/vol – extrapolation done, temp correction
Temp correction – using heating/cooling curve
Thermometric Titration
initial Temp
= 21 C
final Temp
= 29 C
Excel plot
Heating curve
y = 0.31 x + 21.1
Cooling curve
y = - 0.104 x + 31.7
Heating curve
y = 0.31 x + 21.1
Cooling curve
y = - 0.104 x + 31.7
Solving for , x and y
y = + 0.31 x + 21.1 y = - 0.104 x + 31.7
+ 0.31 x + 21.1 = -0.104 x + 31.7
x = 26 (vol acid) y = 29 (Temp)
Weak acid with Strong alkali
CH3COOH + NaOH→CH3COONa + H2O ∆H = - 51 kJ mol-1
Strong acid vs Strong alkali
Weak acid vs strong alkali

18. Vol acid = 25 ml
M x V (acid) = M x V (alkali)
M x 25 = 1 x 50
M (acid) = 2 M
Vol/ml 0 5 10 15 20 25 30 35 40 45 50
Temp/C 22 24. 26 28 30 31 30 30 29 28 27
∆H neutralization rxn
HCI + NaOH → NaCI + H2O ∆H = ?
???? M HCI – 5 ml added and temp recorded
∆H rxn = Heat absorb water
= (mc∆θ)
= 75 x 4.18 x (31 – 22)
= 75 x 4.18 x 9
= - 2821J
V, NaOH = 50 ml
M, = 1M
T initial = 22 C
T final = 31 C by extrapolation.
Coffee cup calorimeter Data collection
Vol, acid = 25
Max Temp = 31
∆T = ( 31 – 22) = 9 C
0.05 mol = - 2821 J
1 mol = - 56 kJ mol-1
Amt OH- = MV
used = 1 x 0.05
= 0.05 mol
% Error = (57.3 – 56) x 100% = 2 %
57.3
Lit value = - 57.3 kJ mol -1
Assump valid if rate of cooling constant
If not constant - extrapolation is wrong
Temp correction – using heating/cooling curve
Thermometric Titration
HCI + NaOH → NaCI + H2O
initial Temp
= 22 C
final Temp
= 31 C
Excel plot
Heating curve
y = 0.362x + 22.4
Cooling curve
y = - 0.16 x + 35.7
Heating curve
y = 0.362x + 22.4
Cooling curve
y = - 0.16 x + 35.7
Solving for , x and y
y = + 0.362 x + 22.4 y = - 0.16 x + 35.7
0.362 x + 22.4 = -0.16 x + 35.7
x = 25 (vol acid) y = 31 (Temp)
Strong acid with Strong alkali Thermometric titration – find conc unknown acid
1 mol HCI = 1 mol NaOH
limiting
HCI + NaOH → NaCI + H2O ∆H = - 56 kJ mol -1

19. Time/m 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
Temp/C 28 28 28 28 28 28 28 30 30.2 29 28
25 ml, 0.5M KCI
No heat loss from system
(isolated system)
↓
∆H system = O
Heat lost rxn = Heat absorb solution
= (mc∆θ)
= 50 x 4.18 x (31 – 28)
= 50 x 4.18 x 3
= - 627J
Coffee cup calorimeter Data collection
Extrapolation best curve fit
y = -2.68x + 39
y = -2.68 x 3 + 39
y = 31 (Max Temp)
initial Temp = 28 C
final Temp = 31 C
time, x = 3
using Excel plot
Using data logger
Min gradient
Extrapolation min gradient
y = -2.68x + 39
y = -2.68 x 3 + 39
y = 31 (Min Temp)
Extrapolation max gradient
y = -3.22x + 44
y = -3.22 x 3 + 44
y = 34 (Max Temp)
Using data logger
Max gradient
Ave Max Temp = (34 + 31)/2
= 32 C
0.0125 mol = - 627 J
1 mol = - 50 kJ mol-1
Enthalpy precipitation = 1 mol of AgCI precipitated
Amt Cu 2+ = MV
used = 0.5 x 0.025
= 0.0125 mol
Ag+ + CI- → AgCI ∆H = -50 kJ mol -1
% Error = (65 – 50) x 100% = 23 %
65
Lit value = - 65 kJ mol -1
Assump valid if rate of cooling constant
If not constant - extrapolation is wrong
Rxn slow – lose heat to surrounding
Plot Temp/time – extrapolation done, Temp correction
Temp correction – using cooling curve for last 5 mins
∆H precipitation rxn
AgNO3 + KCI → AgCI + KNO3
V, AgNO3 = 25 ml
M, = 0.5 M
T initial = 28 C
T final = 31 C
Measure temp AgNO3,
every 0.5m interval
then add KCI (stirring)
Using excel plot ∆T = (31 - 28) = 3 C
Using max/min plot ∆T = (32 - 28) = 4 C
% Error = (65 – 66) x 100% = 2 %
65
CONTINUE
Min/Max gradient

20. Time/m 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
Temp/C 28 28 28 28 28 28 28 30 30.2 29 28
25 ml, 0.5M KCI
No heat loss from system
(isolated system)
↓
∆H system = O
Heat lost rxn = Heat absorb solution
= (mc∆θ)
= 50 x 4.18 x (31 – 28)
= 50 x 4.18 x 3
= - 627J
Coffee cup calorimeter Data collection
Extrapolation best curve fit
y = -2.68x + 39
y = -2.68 x 3 + 39
y = 31 (Max Temp)
initial Temp = 28 C
final Temp = 31 C
time, x = 3
using Excel plot
0.0125 mol = - 627 J
1 mol = - 50 kJ mol-1
Enthalpy precipitation = 1 mol of AgCI precipitated
Amt Cu 2+ = MV
used = 0.5 x 0.025
= 0.0125 mol
Rxn slow – lose heat to surrounding
Plot Temp/time – extrapolation done, Temp correction
Temp correction – using cooling curve for last 5 mins
∆H precipitation rxn
AgNO3 + KCI → AgCI + KNO3
V, AgNO3 = 25 ml
M, = 0.5 M
T initial = 28 C
T final = 31 C
Measure temp AgNO3,
every 0.5m interval
then add KCI (stirring)
Silver
Halides
Enthalpy
Precipitation
AgCI -65 kJ mol-1
AgBr - 85 kJ mol-1
AgI - 111 kJ mol-1
Enthalpy precipitation AgCI lowest
↓
Size CI- – small– strong lattice enthalpy– hydrated by H2O
↓
Lot energy need to break to release CI-
↓
Free CI- will precipitate with Ag+ form AgCI
Anion Enthalpy
Hydration
CI- -359 kJ mol-1
Br- - 328 kJ mol-1
I- - 287 kJ mol-1
Smaller size--↑ -ve ∆H hydration
More heat released (-ve)
CI- Br- I-
CI-
Br-
I-
Ag+ + CI- → AgCI ∆H = -50 kJ mol -1
Bigger size--↑ - Weaker attraction bet I- with H2O
↓
Easier to release I- to form ppt
∆H hydration

21. C2H5OH + 3O2 → 2CO2 + 3H2O
Cold water = 50g
Mass cold water add = 50 g
Mass warm water add = 50 g
Initial Temp flask/cold water = 23.1C
Initial Temp warm water = 41.3C
Final Temp flask/mixture = 31C
1. Find heat capacity calorimeter
Ti = 23.1C
Hot water = 50g
T i = 41.3C
No heat loss from system
(isolated system)
↓
∆H system = O
Heat lost hot H2O = Heat absorb cold H2O + Heat absorb flask
(mc∆θ) (mc∆θ) (c ∆θ)
50 x 4.18 x (41.3 – 31) = 50 x 4.18 x (31-23.1) + c x (31-23.1)
Heat capacity flask, c = 63.5JK-1
∆Hrxn = Heat absorb water + flask
(mc∆θ) + (c∆θ)
∆Hrxn= 250 x 4.18 x (58 – 30) + 63.5 x (58 – 30)
∆Hrxn= - 31 kJ
Lit value = - 1368 kJ mol -1
∆Hc combustion using calorimeter
Mass water = 250 g
Initial Temp water = 30 C
Final Temp water/flask = 58 C
2. Find ∆Hc combustion data
Mass initial spirit lamp/alcohol = 218.0 g
Mass final spirit lamp/alcohol = 216.6 g
Mass alcohol combusted = 1.4 g
0.03 mol = - 31 kJ
1 mol = - 1033 kJ mol-1
Error Analysis
Heat loss to surrounding
Heat capacity sol not 4.18 JK -1
Incomplete combustion (black soot)
Evaporation alcohol
Distance too far
Water not stir
(Heat distribution uneven)
Use bomb calorimeter
% Error = (1368 – 1033) x 100% = 24 %
1368
Mol alcohol = 1.4/46
= 0.03 mol
Hydrocarbon ∆Hc
kJ mol-1
CH3OH - 726
C2H5OH - 1368
C3H7OH - 2021
C4H9OH - 2676
C5H11OH - 3331
Longer hydrocarbon chain
↓
∆H combustion more -ve
C- OH
C-C- OH
C-C-C- OH
C-C-C-C- OH
C-C-C-C-C- OH
Hydrocarbon ∆Hc
kJ mol-1
CH3OH - 726
C2H5OH - 1368
C3H7OH - 2021
C4H9OH - 2676
C5H11OH -3331
H H H H
ᴵ ᴵ ᴵ ᴵ
H – C – C – C – C - OH
ᴵ ᴵ ᴵ ᴵ
H H H H
H H OH H
ᴵ ᴵ ᴵ ᴵ
H – C– C – C – C -H
ᴵ ᴵ ᴵ ᴵ
H H H H
OH
ᴵ
CH3 – C – CH3
ᴵ
CH3
Will structural isomer have same ∆Hc
∆H = -1676kJ mol-1
Research Question ?
О
О
О

22. Time/m 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
Temp/C 25 25 25 25 25 25 25 29 29.4 29 28
0.848 g LiCI
No heat loss from system
(isolated system)
↓
∆H system = O
Heat lost rxn = Heat absorb solution
= (mc∆θ)
= 36 x 4.18 x (29.8 – 25)
= 36 x 4.18 x 4.8
= - 0.72 kJ
Coffee cup calorimeter Data collection
Extrapolation best curve fit
y = -2.68x + 38
y = -2.68 x 3 + 38
y = 29.8 (Temp)
initial Temp = 25 C
final Temp = 29.8 C
time, x = 3
using Excel plot
0.02 mol = - 0.72 kJ
1 mol = - 36 kJ mol-1
Amt LiCI = 0.848/42.4
used = 0.02 mol
LiCI(s) + H2O → LiCI(aq) ∆H = -36 kJ mol -1
Rxn slow – lose heat to surrounding
Plot Temp/time – extrapolation done, Temp correction
Temp correction – using cooling curve for last 5 mins
∆H solution rxn
LiCI (s) + H2O → LiCI (aq)
Mass water = 36 ml
T initial = 25 C
T final = 29.8 C
Lithium
salts
Enthalpy
solution
LiF +4.7 kJ mol-1
LiCI -37 kJ mol-1
LiBr - 48 kJ mol-1
LiI - 63 kJ mol-1
∆H solution
CI-
Br-
I-
limiting
F-
Lithium salt Lattice enthalpy
kJ mol-1
LiF + 1050
LiCI + 864
LiBr + 820
LiI + 764
∆H(sol) = ∆HLE + ∆Hhyd(cation) + ∆Hhyd(anion)
∆Hlattice ∆Hhydration
Size anion ↑
depends
Cation Enthalpy Hyd
kJ mol-1
Li+ - 538
Anion Enthalpy Hyd
kJ mol-1
F- - 504
CI- -359
Br- - 328
I- - 287CONTINUE

23. 0.848 g LiCI No heat loss from system
(isolated system)
↓
∆H system = O
Heat lost rxn = Heat absorb solution
= (mc∆θ)
= 36 x 4.18 x (29.8 – 25)
= 36 x 4.18 x 4.8
= - 0.72 kJ
0.02 mol = - 0.72 kJ
1 mol = - 36 kJ mol-1
Amt LiCI = 0.848/42.4
used = 0.02 mol
LiCI(s) + H2O → LiCI(aq) ∆H = -36 kJ mol -1
∆H solution rxn
LiCI (s) + H2O → LiCI (aq)
Mass water = 36 ml
T initial = 25 C
T final = 29.8 C
Lithium
salt
Enthalpy
solution
kJ mol-1
LiF + 4.7
LiCI - 37
LiBr - 48
LiI - 63
Smaller size -↑ -ve ∆H hyd
More heat released (-ve)
CI- Br-
I-
CI-
Br-
I-
limiting
F-
Std ∆H sol = 1 mol solute dissolved form infinitely dilute sol
∆Hsolution
∆Hlattice ∆Hhydration
∆H(sol) = ∆HLE + ∆Hhyd(cation) + ∆Hhyd(anion)
depends
Lithium
salt
Lattice enthalpy
kJ mol-1
LiF + 1050
LiCI + 864
LiBr + 820
LiI + 764
Anion Enthalpy Hyd
kJ mol-1
F- - 504
CI- -359
Br- - 328
I- - 287
Size anion ↑
↓
∆H lattice decrease ↓ (less +ve)
↓
∆H sol = ∆H latt + ∆H hyd
∆H sol (–ve) = ∆H hyd > ∆H latt
↓
∆H = -ve (more soluble)
∆H latt > ∆H hyd = +ve ∆H (Insoluble)
∆H hyd > ∆H latt = -ve ∆H (Soluble)
Soluble Insoluble
- ve
energy
Coffee cup calorimeter
Cation Enthalpy Hyd
kJ mol-1
Li+ - 538
∆H hydration∆H lattice
+ ve absorb
to break bonds
- ve released
to make bonds
Lattice (+)
Hydration (-)
Hydration > lattice
(-ve ∆H)
Lattice > Hydration
(+ve ∆H)

24. Acknowledgements
Thanks to source of pictures and video used in this presentation
Thanks to Creative Commons for excellent contribution on licenses
http://creativecommons.org/licenses/
Prepared by Lawrence Kok
Check out more video tutorials from my site and hope you enjoy this tutorial
http://lawrencekok.blogspot.com